Question

A man throws a ball into the air with a velocity of 96 ft/s. Use the formula h=−16 t 2 + v 0 t to determine when the height of the ball will be 48 feet. Round to the nearest tenth.

Answer 1

Assuming that v0 represents the initial velocity, then v0 = 96 ft/s. Substitute this value into h(t) and set the equation equal to 48. Then, write this quadratic equation in standard form, which is ax² + bx+ c = 0, where a, b, and c are constants. Either use factoring or the Quadratic Function to solve the equation for t. Remember that t must be positive, because it represents a unit of time.

Answer 2

t=5.4 and t=0.6.

Explanation:

The height of the ball (in feet) after time t is defined by the function

`h=-16t^2+v_0t`

where,
`v_0` is initial velocity.

It is given that a man throws a ball into the air with a velocity of 96 ft/s.

Substitute v=96 in the above function.

`h=-16t^2+96t`

We need to find the time at which the height of ball is 48 feet.

`48=-16t^2+96t`

`16t^2-96t+48=0` .... (1)

Quadratic formula for
`ax^2+bx+c=0` is

`x=(-b\pm √(b^2-4ac))/(2a)`

In equation a=16, b=-96 and c=48. Using quadratic formula we get

`t=(-(-96)\pm √((-96)^2-4(16)(48)))/(2(16))`

`t=(96\pm √(9216-3072))/(32)`

`t=(96\pm 78.384)/(32)`

`t=(96+78.384)/(32)` and
`t=(96-78.384)/(32)`

`t=5.449` and
`t=0.55`

Round the answer to the nearest tenth.

`t\approx 5.4` and
`t\approx 0.6`

Therefore, the height of ball will be 48 feet at t=5.4 and t=0.6.